Problem: The equation of a circle $C$ is $(x-2)^{2}+(y+6)^{2} = 4$. What are its center $(h, k)$ and its radius $r$ ?
Solution: The equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$ We can rewrite the given equation as $(x - 2)^2 + (y - (-6))^2 = 2^2$ Thus, $(h, k) = (2, -6)$ and $r = 2$.